續二次方程(複數)(27-30題)

27a) 5-i/(1+i)(2+3i)
= 5-i /(2+ 3i + 2i - 3) * i x i = -1*
= 5-i /(5i-1)
=(5-1)(5i+1) / (-25-1)
=(10+24i) / (-26)
=-5/13 - 12i/13
b) (3+i)(2-i)/4+3i
= (6 -3i + 2i +1) / (4+3i)
= (7- i) / (4+3i)
= (7-i)(4-3i) / (16-9)
=(28- 21i -4i -3)/ 7
=25/7 - 25i/7

28a)(√3 +i)(√3 -i)
= 9-(-1)
=10
b)(√3 + i)^3 (√3 -i)^2
=10 x 10 x (3^(1/2)+i)
=100(3)^(1/2) + 100i

29a)2i- 4/3-i
= (2i-4)(3+i) / (9+1)
= (6i -2 -12 -4i) / 10
= (2i-14)/10
= -7/5 + i/5
29b)(2+5i)^2 + (8-10i)
= 4 + 20i -25 + 8-10i
=10i-13

30a)2i(-4+3i)+(4-5i)(4+5i)
= -8i -6 +16 +25
=-8i-3
30b. (4-2i)^2 - (1+i)^2
=16 +16i +4 -1 -2i +1
=14i+12

27a) (5-i)/[(1+i)(2+3i)]
= (5 - i)/(-1 + 5i)
= (5 - i)(-1 - 5i)/26
= -(5/13) - (12/13)i

27b) (3+i)(2-i)/(4+3i)
= (7 - i)/(4 + 3i)
= (7 - i)(4 - 3i)/25
= 1 - i

28a) (根3+i)(根3-i) = 10

28b) (根3+i)^3 (根3-i)^2
= 10^2 (根3 + i)
= 100根3 + 100i

29a) (2i-4)/(3-i)
= 2(-2 + i)(3 + i)/10
= (-7 + i)/5
= -(7/5) + (1/5)i

29b) (2+5i)^2 + (8-10i)
= 4 + 20i - 25 + 8 - 10i
= -13 + 10i

30a) 2i(-4+3i) + (4-5i)(4+5i)
= -6 - 8i + 16 + 25
= 35 - 8i

30b) (4-2i)^2 - (1+i)^2
= (16 - 16i - 4) - (1 + 2i - 1)
= 12 - 18i